Question

If $y=(\log x{)}^x+x^{\log x}$, find $\frac{dy}{dx}$.

Answer 1

Given $y={\left(\log x\right)}^x+x^{\log x}$

Let $h={\left(\log x\right)}^x$

Apply $\log$ on both sides

$\log h=x\log \left(\log x\right)$

Differentiate on both sides w.r.t. x

$\frac{1}{h}\frac{dh}{dx}=\log \left(\log x\right)+x.\frac{1}{\log x}.\frac{1}{x}$

$\therefore \frac{dh}{dx}={\left(\log x\right)}^x[\log \left(\log x\right)+\frac{1}{\log x}]$

Let $t=x^{\log x}$

Apply $\log$ on both sides

$\log t=\log x\log x={\left(\log x\right)}^2$

Differentiate on both sides w.r.t. x

$\frac{1}{t}.\frac{dt}{dx}=2\left(\log x\right).\frac{1}{x}$

$\therefore \frac{dt}{dx}=2x^{\log x}\left(\frac{\log x}{x}\right)$

$y=h+t$

$\frac{dy}{dx}=\frac{dh}{dx}+\frac{dt}{dx}$

$\therefore \frac{dy}{dx}={\left(\log x\right)}^x[\log \left(\log x\right)+\frac{1}{\log x}]+2x^{\log x}.\left(\frac{\log x}{x}\right)$